Question: Solve for $a$, $ -\dfrac{8}{8a^2} = \dfrac{4a - 3}{4a^2} + \dfrac{3}{20a^2} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8a^2$ $4a^2$ and $20a^2$ The common denominator is $40a^2$ To get $40a^2$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{8}{8a^2} \times \dfrac{5}{5} = -\dfrac{40}{40a^2} $ To get $40a^2$ in the denominator of the second term, multiply it by $\frac{10}{10}$ $ \dfrac{4a - 3}{4a^2} \times \dfrac{10}{10} = \dfrac{40a - 30}{40a^2} $ To get $40a^2$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{3}{20a^2} \times \dfrac{2}{2} = \dfrac{6}{40a^2} $ This give us: $ -\dfrac{40}{40a^2} = \dfrac{40a - 30}{40a^2} + \dfrac{6}{40a^2} $ If we multiply both sides of the equation by $40a^2$ , we get: $ -40 = 40a - 30 + 6$ $ -40 = 40a - 24$ $ -16 = 40a $ $ a = -\dfrac{2}{5}$